For some reason I can\'t get this problem right. A dart gun is fired while being
ID: 1750931 • Letter: F
Question
For some reason I can't get this problem right. A dart gun is fired while being held horizontally at a heightof 1.50 m above ground level and while itis at rest relative to the ground. The dart from the gun travels ahorizontal distance of 4.50 m. A collegestudent holds the same gun in a horizontal position while slidingdown a 45.0° incline at a constant speed of 2.00 m/s. How far will the dart travel if the studentfires the gun when it is 1.00 m above the ground?For some reason I can't get this problem right. A dart gun is fired while being held horizontally at a heightof 1.50 m above ground level and while itis at rest relative to the ground. The dart from the gun travels ahorizontal distance of 4.50 m. A collegestudent holds the same gun in a horizontal position while slidingdown a 45.0° incline at a constant speed of 2.00 m/s. How far will the dart travel if the studentfires the gun when it is 1.00 m above the ground?
Explanation / Answer
the 1st part of the question allows you to find the velocity of adart fired from the gun * the dart travels 4.50 m in the time it takes to fall 1.50 m * d = (1/2)gt2 ___ 1.50 = .5 * 9.81 *t2 ___ t = .553 s * so the fired velocity is 4.50 / .553 or 8.14m/s * * * on the slide, the horizontal component of the sliding velocity isadded to the fired velocity and the vertical component gives the falling dart an initialdownward velocity which shortens the "flight" time * because the sliding angle is 45º, the horizontal and verticalcomponents are equal 2.00 * sin(45º) = 1.41 m/s * the horizontal velocity of the dart from the slide is 8.14 +1.41 or 9.55 m/s * to find the "flight" time ___ d = .5gt2 +v0t ___ 1.00 = 4.9t^2 + 1.41t ___ 0 = 4.9t2 + 1.41t - 1.00 using quadratic formula ___ t = .330 s * the distance the slide-launched dart travels is 9.55 *.330 or 3.15 m